How best to estimate the power losses when harmonics are attenuated from a square wav

[JAMES IRVINE]

I am currently doing a project for a dc to ac inverter for my hnd graded unit.
I must submit a calculation and was wondering if anyone can tell me if my idea has any credibility if not has anyone any ideas.
My idea is to estimate the power losses which will result from using a low pass filter to attenuate unwanted harmonics from a squarewave output to produce a sinewave.
This is the link to the circuit which I intend to use **broken link removed**

I have to achieve a 50 HZ 240v output not 230 for some odd reason and I calculate that the current through the primary will be 30A, which I have assumed means that the load impedance must be 10hm: . I dont know if this is acceptable or whether I need to use thevenims theorem to find out the circuit impedance.
I am thinking that if I calculate using thevenims theorem to find the circuit resitance, inductance and capacitance then I can determine the power losses using the various equations for complex waves like v=16 sin wt + etc.
The problem I see with that is determining the value of the peak voltage for the 3rd harmonic,5th......, I mean can I calculate that or would it have to be measured.
For the low pass filter values I have took a wild guess that they have to resonate at 50Hz which I guess means I use a value of j35 to substitute into equations for XL and XC, I'm unsure if the resistance of the load must be taken into
consideration or whether I am miles away from the correct theory on this.
After all I know that resistance will have a damping effect and may mean that resonance would occur at different value of L and C.

 

[Nigel Goodwin]

Are you perhaps overcomplicating it?.

Essentially you're starting with a squarewave of a certain amplitude, presumably you can easily calcuate the amount of power supplied into a specific load?.

Now low-pass filter that to give a sinewave, assume zero losses so you have a sinewave of the same peak to peak value as the squarewave, then calculate the power into the same load as before.

Power loss is first result minus second result.

 

[Roff]

Are you perhaps overcomplicating it?.

Essentially you're starting with a squarewave of a certain amplitude, presumably you can easily calcuate the amount of power supplied into a specific load?.

Now low-pass filter that to give a sinewave, assume zero losses so you have a sinewave of the same peak to peak value as the squarewave, then calculate the power into the same load as before.

Power loss is first result minus second result.

Just a minor nitpick- The peak-to-peak amplitude of the fundamental is 4/pi (~1.27) times the peak-to-peak amplitude of the square wave.
https://cnyack.homestead.com/files/afourse/fssq1.htm

 

[JAMES IRVINE]

I have no real experience in this field so none of this means anything but what about the legitamacy of my idea for creating a calculation for my project. Is it possible to predict the voltage or current value of the third harmonic.

 

[pc88]

You could always simulate your circuit using something like ltspice: https://www.linear.com/designtools/software/

 

[audioguru]

An LC circuit tuned to 50Hz will have such high values that the resistance of the inductor will reduce the output power to nearly nothing.

If you have a lowpass filter without loss, and the output is a perfect square-wave (it is not, look at the 'scope photos in the project) then the output power in the sinewave is exactly half the output power of the square-wave.

Without a load then the LC lowpass filter will be a series resonant circuit which is a dead short.